Question

The square root of $-1+2\sqrt{2}i$ is

• A. $\pm (2+\sqrt{2}i)$
• B. $\pm (3+\sqrt{2}i)$
• C. $\pm (-1+\sqrt{2}i)$
• D. $\pm (1+\sqrt{2}i)$

Answer 1

The correct option is C $\pm (-1+\sqrt{2}i)$

Let $\sqrt{-1+2\sqrt{2}i}$ = a + ib

square on both sides

$-1+2\sqrt{2}i=a^2-b^2+2abi$

Comparing real and imaginary part on both sides

$a^2-b^2=-1\dots \dots$ (1)

$2abi=2\sqrt{2}i\dots \dots$ (2)
$\Rightarrow ab=\sqrt{2}$

We know,

$a^2+b^2=\sqrt{(a^2-b^2{)}^2+4a^2{b}^2}$

$a^2+b^2=\sqrt{1+8}=\sqrt{9}=\pm 3$

$a^2+b^2=3\dots \dots$ (3)

$a^2-b^2=-1\dots \dots$ (4)

$2a^2=2$

$a^2=1,a=\pm 1$

Substitute a in equation 3

$b^2=3-1=2$

$b=\pm \sqrt{2}$

The real part of $-1+i\sqrt[2]{22}$ is negative. Hence, the real part of the square root is negative as well

Since imaginary part of $-1+i\sqrt[2]{22}$ is positive, imaginary part of square root$\pm (1+i\sqrt{2})$ should also be positive.

Square root is $\pm (-1+\sqrt{2}i)$